Question: Find the focus of the parabola $x = -\frac{1}{12} y^2.$
Explanation: Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix.

Since the parabola $x = -\frac{1}{12} y^2$ is symmetric about the $x$-axis, the focus is at a point of the form $(f,0).$  Let $x = d$ be the equation of the directrix.

[asy]
unitsize(1.5 cm);

pair F, P, Q;

F = (-1/4,0);
P = (-1,1);
Q = (-1/4,1);

real parab (real x) {
  return(-x^2);
}

draw(reflect((0,0),(1,1))*graph(parab,-1.5,1.5),red);
draw((1/4,-1.5)--(1/4,1.5),dashed);
draw(P--F);
draw(P--Q);

dot("$F$", F, SW);
dot("$P$", P, N);
dot("$Q$", Q, E);
[/asy]

Let $\left( -\frac{1}{12} y^2, y \right)$ be a point on the parabola $x = -\frac{1}{12} y^2.$  Then
\[PF^2 = \left( -\frac{1}{12} y^2 - f \right)^2 + y^2\]and $PQ^2 = \left( -\frac{1}{12} y^2 - d \right)^2.$  Thus,
\[\left( -\frac{1}{12} y^2 - f \right)^2 + y^2 = \left( -\frac{1}{12} y^2 - d \right)^2.\]Expanding, we get
\[\frac{1}{144} y^4 + \frac{f}{6} y^2 + f^2 + y^2 = \frac{1}{144} y^4 + \frac{d}{6} y^2 + d^2.\]Matching coefficients, we get
\begin{align*}
\frac{f}{6} + 1 &= \frac{d}{6}, \\
f^2 &= d^2.
\end{align*}From the first equation, $d - f = 6.$  Since $f^2 = d^2,$ $f = d$ or $f = -d.$  We cannot have $f = d,$ so $f = -d.$  Then $-2f = 6,$ so $f = -3.$

Thus, the focus $\boxed{(-3,0)}.$